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A space $X$ is a Urysohn space iff whenever $x \neq y$ in $X$ there are nhoods of $U$ of $x$ and $V$ of $y$ such that $\overline{U} \cap \overline{V} = \emptyset$.

I want to show that every regular, T_1 space is Urysohn.

My attempt:

Let $x,y \in X$ be distinct points. Now consider the open set $X \setminus \{y\}$. By regularity we can find an open set $U$ such that $x \in U \subseteq \overline{U} \subseteq X \setminus \{y\}$. Now similarly by considering the open set $X \setminus \{x\}$ and using regularity again we can find an open set $V$ such that $y \in V \subseteq \overline{V} \subseteq X \setminus \{x\}$.

Now from where I don't think we can conclude $U$ and $V$ have disjoint closures. Can we? I'm stuck here, can you please help?

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+1 for showing what you have done – Ross Millikan Jan 6 '11 at 2:31

1 Answer 1

Suppose $X$ is $T_1$ and regular.

Take $x,y\in X$ two different points. Consider, as you did, $X\setminus\{y\}$ as an open neighbourhood of $x$, there exists $x\in U\subseteq\bar{U}\subseteq X\setminus\{y\}$.

Consider now, instead of your choice, $X\setminus\bar{U}$. It is an open neighbourhood of $y$ and from regularity you have $y\in V\subseteq\bar{V}\subseteq X\setminus\bar{U}$.

$\bar{U}\cap\bar{V}=\emptyset$ as needed.

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Very nice, thank you! – user Jan 6 '11 at 3:55

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